 Solution

An intrinsic semiconductor has negligible electrically active impurity doping, and thermal excitation across its band gap produces equal electron and hole densities $n=p=n_i$. In an extrinsic sample, semiconductor doping supplies donor impurities or acceptor impurities, so one carrier species dominates over a range of temperatures. Doping does not remove the possibility of low-temperature dopant freeze-out or high-temperature intrinsic behavior.

Assume parabolic isotropic band edges, spin degeneracy two, a single valley, thermal equilibrium and nondegenerate carriers. Write $E_F$ for the Fermi level, and measure conduction-band energy by $E=E_c+ϵ$. Counting momentum states gives the density of states per unit volumeThe low-occupation Maxwell–Boltzmann statistics givesThe substitution $x=ϵ/(k_{B}T)$ and ${\int }_0^{\infty }\sqrt{x}{e}^{-x}dx=\sqrt{\pi }/2$ give the coefficient. Similarly $p=N_v{e}^{(E_v-E_F)/(k_{B}T)}$ and $N_v=2[m_h^{\ast }{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$. Multiplication eliminates $E_F$, proving the semiconductor mass-action lawHere the supplied effective masses are treated as the masses entering the effective density of states. A detailed multivalley silicon model would require the appropriate valley degeneracy and density-of-states masses consistently; adding an extra factor without changing the stated model would alter the requested expression.

A pentavalent arsenic impurity supplies a weakly bound excess electron. In the hydrogenic donor state approximation, dielectric screening reduces the Coulomb attraction and the band effective mass replaces the free mass. Hence the donor binding energy isThis is the effective-mass estimate, without a material-specific central-cell correction.

For the temperature plot use donor density $N_D=10^{15}{\mathrm{cm}}^{-3}$ and no acceptors. The charge neutrality with partial donor ionization model puts $E_D=E_c-E_b$ and includes the twofold spin degeneracy of an occupied donor:These equations give three regimes. At low temperature, $p$ is negligible and strong dopant freeze-out gives $n\simeq \sqrt{N_D{N}_c/2}{e}^{-E_b/(2k_{B}T)}$. At intermediate temperature, the donors are almost ionized and intrinsic holes remain rare, so $n\simeq N_D$. At high temperature, $n\simeq p\simeq n_i\propto T^{3/2}{e}^{-E_g/(2k_{B}T)}$. Thus the log-density plot against inverse temperature has a steep intrinsic segment on the left, a nearly horizontal extrinsic plateau, and a freeze-out decline on the right; the $T^{3/2}$ and $T^{3/4}$ factors give slight curvature.

To quantify “approximately constant”, require $0.9N_D\le n\le 1.1N_D$. At $300\mathrm{K}$, the specified model gives $N_c\simeq 4.12\times 10^{18}$, $N_v\simeq 8.61\times 10^{18}$ and $n_i\simeq 2.33\times 10^9{\mathrm{cm}}^{-3}$. The full neutrality equations give the ten-percent plateau intervalThe lower boundary is set by donor ionization and the upper boundary by intrinsic electron-hole production. The large ratio $N_c/N_D$ allows substantial donor ionization even when $k_{B}T<E_b$; using only $k_{B}T\gg E_b$ would incorrectly exclude much of the plateau. The quoted interval depends on the chosen tolerance and the constant-gap band model, rather than being an exact measured range for arsenic-doped silicon.